= Solution
Set $\Delta=T/M$, $q_0=q$, and $q_M=q'$. The <Trotter product formula> separates the kinetic and <potential energy> evolution to first order in $\Delta$. Inserting $M-1$ resolutions of the identity in the <position operator> basis and using the free short-time kernel gives
$$
K(q',q;T)=\lim_{M\to\infty}(2\pi i\Delta)^{-M/2}
\int\prod_{j=1}^{M-1}dq_j\,
\exp\left\{i\sum_{j=1}^{M}\left[\frac{(q_j-q_{j-1})^2}{2\Delta}-\Delta V(q_{j-1})\right]\right\}.
$$
The exponent converges formally to the <action> along the path. The prefactors and intermediate integrations define the <time-sliced configuration-space path integral> measure, so
$$
\boxed{K(q',q;T)=\int_{q(0)=q}^{q(T)=q'}\mathcal Dq\,e^{iS[q]}.}
$$
A real-time <path integral> is an oscillatory limit with the <Feynman i-epsilon prescription>, rather than an ordinary probability integral. The operator derivation assumes the usual <self-adjointness> and product-formula hypotheses for the Hamiltonian.
For a quadratic <potential energy> write $q=q_c+\eta$, with $\eta(0)=\eta(T)=0$ and $\ddot q_c+V'(q_c)=0$. The first variation vanishes by the <Euler-Lagrange equation> and the fixed endpoints. Because $V$ is quadratic, the remaining expansion is exact:
$$
S[q_c+\eta]=S[q_c]+\frac12\int_0^T\bigl(\dot\eta^2-V''\eta^2\bigr)dt
=S[q_c]-\frac12\int_0^T\eta A\eta\,dt,
\qquad A=\partial_t^2+V''.
$$
Here <integration by parts> has no boundary contribution. Translation of the integration variables leaves the time-sliced measure unchanged. Thus the <Gaussian path integral> over $\eta$ is independent of the endpoint values, and \b[the entire endpoint dependence lies in $e^{iS[q_c]}$].
Expanding $\eta$ in <eigenfunctions> satisfying <Dirichlet boundary conditions>, each real mode contributes an inverse square root of its quadratic <eigenvalue>, with the oscillatory phase fixed by continuation. Normalizing against $A_0=\partial_t^2$ gives a precise determinant-ratio form:
$$
\boxed{N(T)=\frac1{\sqrt{2\pi iT}}
\left(\frac{\det_D A}{\det_D A_0}\right)^{-1/2}.}
$$
The <determinant> ratio is meaningful after a common regulator; the phase is part of the prescription. It is equivalent to writing $N=C(T)(\det_D A)^{-1/2}$ with an endpoint-independent normalization.
For example, if $V''=\omega^2$, the <Dirichlet oscillator determinant ratio> is
$$
\frac{\det_D A}{\det_D A_0}
=\prod_{r=1}^{\infty}\left(1-\frac{\omega^2T^2}{r^2\pi^2}\right)
=\frac{\sin\omega T}{\omega T},
\qquad N=\left(\frac{\omega}{2\pi i\sin\omega T}\right)^{1/2}.
$$
A linear or constant term in $V$ changes the <action> but not this prefactor. The formula holds away from conjugate times, when the boundary problem and <determinant> are nonsingular. At $\sin\omega T=0$, it must be continued as a <distribution> with its <Maslov index> phase; the assertion of an ordinary finite prefactor cannot be used literally there. For the unshifted oscillator, at $T=r\pi/\omega$ the kernel is $e^{-ir\pi/2}\delta(q'-(-1)^r q)$.
For the full-line source integral, impose vacuum boundary conditions by adiabatic damping. Take the source initially as a <test function>, so its pairings with the <Green function> are defined. After <integration by parts>, write
$$
S[q]=\frac12(q,Kq)+(J,q),\qquad K=-\partial_t^2-m^2+i\epsilon.
$$
With the <Fourier transform> convention $G(t)=\int d\omega\,e^{i\omega t}\widetilde G(\omega)/(2\pi)$, the inverse is
$$
\widetilde G(\omega)=\frac1{\omega^2-m^2+i\epsilon},\qquad KG=\delta.
$$
Set $q=\eta-GJ$. <Completing the square> gives
$$
S[q]=\frac12(\eta,K\eta)-\frac12\int dt\,dt'\,J(t)G(t-t')J(t'),
$$
and consequently
$$
\boxed{Z[J]=Z[0]\exp\left[-\frac i2\int dt\,dt'\,J(t)G(t-t')J(t')\right].}
$$
The <Gaussian functional integral> $Z[0]$ is source independent and contains the regulated <determinant>. This is most naturally read as a <normalized vacuum generating functional>. For $m>0$, closing the frequency contour gives $G(t)=-i e^{-im|t|}/(2m)$; the time-ordered oscillator two-point function is $iG(t)$. This checks the sign of the source exponent and identifies the required vacuum prescription.
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